Calculus

Question

i really don't understand how to go about finding the derivative for this if f(x) = sin^2(3-x) then f '(0) = ? A. -2cos3 B. -2sin3cos3 C. 6cos3 D.2sin3cos3 E. 6sin3cos3

Answer

This function is of the form u(x)^n, whre u(x) = sin(3-x) and n=2. The derivative is n*u(x)^(n-1)*u'(x). For u=sin(x), then u'=cos(x). So in this case the derivative is 2*sin(3-x)*cos(3-x)*(-1) where the -1 at the end comes from derivative of the argument of cos. This is rearranged to = -2sin(3-x)cos(3-x). So f'(0)=-2sin3cos3 or answer B. http://vrd.askvrd.org/services/answerschema.xml